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1.Units, Dimensions and Measurement
medium
From the dimensional consideration, which of the following equation is correct
A$T = 2\pi \sqrt {\frac{{{R^3}}}{{GM}}} $
B$T = 2\pi \sqrt {\frac{{GM}}{{{R^3}}}} $
C$T = 2\pi \sqrt {\frac{{GM}}{{{R^2}}}} $
D$T = 2\pi \sqrt {\frac{{{R^2}}}{{GM}}} $
Solution
(a) By substituting the dimensions in $T = 2\pi \sqrt {\frac{{{R^3}}}{{GM}}} $
we get $\sqrt {\frac{{{L^3}}}{{{M^{ – 1}}{L^3}{T^{ – 2}} \times M}}} = T$
we get $\sqrt {\frac{{{L^3}}}{{{M^{ – 1}}{L^3}{T^{ – 2}} \times M}}} = T$
Standard 11
Physics
Similar Questions
Choose the correct match
| List I | List II |
|---|---|
| $(i)$ Curie | $(A)$ $ML{T^{ – 2}}$ |
| $(ii)$ Light year | $(B)$ $M$ |
| $(iii)$ Dielectric strength | $(C)$ Dimensionless |
| $(iv)$ Atomic weight | $(D)$ $T$ |
| $(v)$ Decibel | $(E)$ $M{L^2}{T^{ – 2}}$ |
| $(F)$ $M{T^{ – 3}}$ | |
| $(G)$ ${T^{ – 1}}$ | |
| $(H)$ $L$ | |
| $(I)$ $ML{T^{ – 3}}{I^{ – 1}}$ | |
| $(J)$ $L{T^{ – 1}}$ |
